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### Author Topic: Sum of torque  (Read 139765 times)

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #330 on: August 01, 2016, 01:03:45 PM »
I use a continuous track and a road wheel. I suppose the road wheel can receive a torque from Fg (there is an axis of rotation of the wheel). The energy lost by the spring is won by the track but the energy win by the torque on the road wheel is greater than the energy lost by the center of the wheel.

Lg3 is higher than 1/sqrt(2) so the sum of energy is not 0 because the center of the wheel receives always F/sqrt(2) because the direction of the force is always the same: pi/4.
« Last Edit: August 01, 2016, 05:07:52 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #331 on: August 02, 2016, 11:27:45 AM »
I suppose the force from the spring constant even the length increases.

The center of the wheel is (0,0)
With an angle of rotation of 0.01 rd

Point P1:
At start (2 , 1)
At final (2.02 , 1)

Point P2:
At start (1 , 0)
At final (1.00995000043 , -0.009999833)

At start the angle of the forces from the spring is atan( (1-0) / (2-1)) = pi/4 rd
At final the angle of the forces from the spring is atan( (1+0.009999833) / (2.02-1.00995000043) ) = 0.7853733291 rd

cos(pi/4) = √2/2
cos(0.7853486586) = 0.7071243415

The difference is 1.75e-5 so the force on the center of the wheel change from √2/2 to √2/2-1.75e-5
And the torque is exactly √2/2+(0.01-1.75e-5)/√2 the sum of energy can't be at 0

« Last Edit: August 02, 2016, 11:27:08 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #332 on: August 03, 2016, 02:44:43 PM »
I calculated with a program and I find a difference. With quadmath it's the same difference.
« Last Edit: August 04, 2016, 12:03:36 AM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #333 on: August 04, 2016, 03:54:49 PM »
I found my error it was in the equation of the distance from the center.

Maybe with friction and forces from the ground with 2 wheels
« Last Edit: August 05, 2016, 12:16:12 AM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #334 on: August 05, 2016, 03:24:26 PM »
Maybe a wheel and a spike full of water under gravity. I can adjust the contact wall/water like I want. The trajectory of the point is in red. I named the wall of the wheel elastic but maybe it is more dynamic, it is possible to imagine a wall theoretical  that can move like the wheel. I just want to benefit the pressure on the wall A for example.

Image m4p7: it's possible to rotate the spike in the same time the wheel rotates and moves. The spike needs a lower energy to rotate than the wheel won.
« Last Edit: August 05, 2016, 11:44:32 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #335 on: August 10, 2016, 04:12:52 PM »
The torque from the liquid is not the same the torque from the solid, the wheel rotates and moves alone:

The law of gravity is 1/d² not 1/d.
« Last Edit: August 10, 2016, 09:23:03 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #336 on: August 11, 2016, 09:00:18 AM »
Radius of the wheel = 0.5 m
Direction of gravity : vertical
Origin of gravitation: 0.5 m below the ground
Law of attraction: 1/d²

Torque on the solid:

integrate from 0 to 0.5 of integrate from -pi/2 to pi/2 of x^2*cos(y)/(1+x*sin(y))² dx = 0.0986123 Nm

Torque on the walls from the liquid:

integrate from 1.5 to 1 of (x-1)*(1/1.5-1/x) - integrate from 1 to 0.5 of (x-1)*(1/1.5-1/x) dx = 0.0112016 - 0.1098 = - 0.0986123 Nm

There is no torque on the wheel.

Now, if I attract from a point below the red dot at 0.5 m, the pressure on the walls from the liquid don't change but the torque on the liquid change

Like it is the same density for the liquid and the solid, the potential energy of gravity don't lost any energy when the wheel rotates.

« Last Edit: August 11, 2016, 12:17:16 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #337 on: August 11, 2016, 11:43:54 PM »
Like the attraction come from a point (the green point), I need to give an energy in the center of the wheel but that energy is a potential energy that the wheel wins. The torque 0.0986123 - 0.083333 = 0.0152 Nm is the torque that can create an energy when the wheel rotates counterclockwise. Maybe it's possible to use a sphere.

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #338 on: August 12, 2016, 02:28:41 PM »
Like that the cg of the semi disk solid don't work. There is negative torque on the wheel from the liquid on the walls but the work is :

( 0.1098 - 0.0112 ) x = 0.0986 x with x a small angle of rotation

There is an energy from the center of the wheel:

(0.1098 + 0.0112 ) = 0.121 x

The vertical force don't work and the cg of the semi solid disk can't work
« Last Edit: August 12, 2016, 11:00:07 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #339 on: August 13, 2016, 09:03:20 AM »
I have the good values for the forces with the integrals I have the sum of forces at 0 (not with my programs).

The torque from the solid part is -0.252879+0.0675103+1.36114 = 1.5465087 Nm

The torque from the liquid is 1.38147-0.0277572 = 1.3537128 Nm

There is a difference.
« Last Edit: August 13, 2016, 06:59:29 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #340 on: August 13, 2016, 02:39:21 PM »
The wheel lost an energy because the force on the white object is perpendicular of the trajectory, it is possible to controlled the velocity by an external device like that there is no forces on the balls from the white object. When the wheel rotates and moves the white object lost a potential energy.
« Last Edit: August 13, 2016, 06:59:18 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #341 on: August 13, 2016, 07:25:39 PM »
In fact, like I drawn before the spring, the poitn don't work but the spring increases its length !!

The spring is attached between the wheel and the ground. Its length pass from 1.414 R to 1.57 R

Cost nothing for the wheel because the positive clockwise torque from the force is cancelled by the force on the center of the wheel.

The wheel rotates and moves before I attached the spring. There is an inertia, so the wheel continue to move and rotate.
« Last Edit: August 13, 2016, 10:03:40 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #342 on: August 14, 2016, 11:20:05 AM »
I can use a water+gravity device. There is no force on the green point so I don't need any energy to rotate the black arm. There is a torque on the pink arm and like the angle between the black arm and the pink arm change I can recover an energy not with a motor because in this case the motor will give a torque on the black torque, but simply rotate a disk, the disk is on the green axis without friction, it can't give any torque on the black arm. I win the rotation of the disk, potential energy in the rotation of the disk.
« Last Edit: August 14, 2016, 11:00:27 PM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #343 on: August 14, 2016, 11:50:39 PM »
Use a local gravity, replace water by small balls and gravity by springs attached on the yellow point. Turn all the device clockwise. The yellow point turn, it is in the device, the springs to have gravity don't lost any energy.
« Last Edit: August 15, 2016, 11:05:41 AM by EOW »

#### EOW

• Sr. Member
• Posts: 383
##### Re: Sum of torque
« Reply #344 on: August 15, 2016, 04:57:32 PM »
The law of attraction need to be linear in the contrary there is a torque on the container I think.

It's easier to think with a line of attraction like that it's possible to use the formulas of buoyancy/pressure in water with gravity.

I need to adjust the law of attraction in the same time the radius of the black arm change, like that I can have no torque on the container and have one to the pink arm. Change the law of attraction cost nothing in energy because the balls are always in the same position in the container so the springs don't lost any energy.

An example to calculate the law of attraction:
law in 1/d^0.2 with big length for the black arm
« Last Edit: August 15, 2016, 10:52:46 PM by EOW »